Giải thích các bước giải:
\(\begin{array}{l}
2K + 2{H_2}O \to 2KOH + {H_2}\\
2KOH + C{\rm{uS}}{O_4} \to {K_2}S{O_4} + Cu{(OH)_2}\\
{n_K} = 0,2mol\\
{m_{{\rm{CuS}}{O_4}}} = \dfrac{{402,2 \times 8\% }}{{100\% }} = 32,2g\\
\to {n_{{\rm{CuS}}{O_4}}} = 0,2mol
\end{array}\)
-> Phản ứng xảy ra vừa đủ, khôngcó chất dư
\(\begin{array}{l}
1)\\
\to {n_{{H_2}}} = \dfrac{1}{2}{n_K} = 0,1mol\\
\to {V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
{n_{KOH}} = {n_K} = 0,2mol\\
{n_{Cu{{(OH)}_2}}} = \dfrac{1}{2}{n_{KOH}} = 0,1mol\\
\to {m_{Cu{{(OH)}_2}}} = 0,1 \times 98 = 9,8g\\
2)\\
{n_{{K_2}S{O_4}}} = \dfrac{1}{2}{n_{KOH}} = 0,1mol\\
\to {m_{{K_2}S{O_4}}} = 0,1 \times 174 = 17,4g\\
\to {m_{{\rm{dd}}{K_2}S{O_4}}} = {m_{KOH}} + {m_{{\rm{ddCuS}}{O_4}}} - {m_{Cu{{(OH)}_2}}} = 0,2 \times 56 + 402,2 - 9,8 = 403,6g\\
\to C{\% _{{\rm{dd}}{K_2}S{O_4}}} = \dfrac{{17,4}}{{403,6}} \times 100\% = 4,31\% \\
\end{array}\)
